26考研武忠祥17堂课课后习题解析

今天给大家整理出的26重点资源是 👇

•26数学-武忠祥17堂课课后习题解析

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17堂课核心专题与习题重点：专题划分与高频考点极限计算：重点掌握泰勒展开、等价无穷小替换及洛必达法则的适用条件，习题中常出现含参量或嵌套结构的极限问题。中值定理证明：罗尔定理与拉格朗日中值定理的应用需结合构造辅助函数的技巧，习题多考察“双中值”或与不等式结合的综合题。二重积分：极坐标与直角坐标转换是重点，习题常涉及对称性简化计算或分段积分技巧。

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📚高数极限计算例题解析

1. 【解】(limlimits_{n to infty} frac{n}{ln(n + 1)} (sqrt[n]{n} – 1) = limlimits_{n to infty} frac{n}{ln(n + 1)} left( e^{frac{1}{n}ln n} – 1 right) = limlimits_{n to infty} frac{n}{ln(n + 1)} cdot frac{1}{n}ln n = 1)

2. 【解】(e = limlimits_{x to 0} left( frac{1 – tan x}{1 + tan x} right)^{frac{1}{sin kx}} = e^{limlimits_{x to 0} frac{1}{sin kx} ln frac{1 – tan x}{1 + tan x}})指数相等，得 (1 = limlimits_{x to 0} frac{1}{sin kx} cdot frac{-2tan x}{1 + tan x} = limlimits_{x to 0} frac{-2tan x}{kx} = frac{-2}{k})因此 (k = -2)。

3. 【解】原式 (= limlimits_{x to infty} e^{frac{xln x}{(x – 1)(x – 2) cdots (x – n)}} = e^{limlimits_{x to infty} x[ln x – ln(x – 1) – ln(x – 2) – cdots – ln(x – n)]})(= e^{limlimits_{x to infty} xleft[ lnleft(1 + frac{1}{x – 1}right) + lnleft(1 + frac{2}{x – 2}right) + cdots + lnleft(1 + frac{n}{x – n}right) right]})（n项相加）(= e^{limlimits_{x to infty} xleft( frac{1}{x} + frac{2}{x} + cdots + frac{n}{x} right)} = e^{frac{n(n + 1)}{2}})

4. 【解】由(limlimits_{x to +infty} left( frac{x^2 + 3}{x – 2} – ax + b right) = limlimits_{x to +infty} frac{x^2 + 3 – ax^2 + 2ax + bx – 2b}{x – 2})(= limlimits_{x to +infty} frac{(1 – a)x^2 + (2a + b)x – 2b + 3}{x – 2} = 0)可知 (1 – a = 0)，(2a + b = 0)，则 (a = 1)，(b = -2)。答案选 (D)

5. 【解】(limlimits_{x to +infty} frac{x^{6x}}{(x + 1)^x(x + 2)^{2x}(x + 3)^{3x}} = limlimits_{x to +infty} frac{e^{6xln x}}{e^{xln(x + 1)} cdot e^{2xln(x + 2)} cdot e^{3xln(x + 3)}})(= limlimits_{x to +infty} e^{6xln x – [xln(x + 1) + 2xln(x + 2) + 3xln(x + 3)]} = limlimits_{x to +infty} e^{6xln x – xleft[ ln x lnleft(1 + frac{1}{x}right) + 2ln x lnleft(1 + frac{2}{x}right) + 3ln x lnleft(1 + frac{3}{x}right) right]})(= e^{limlimits_{x to +infty} frac{lnleft(1 + frac{1}{x}right) + 2lnleft(1 + frac{2}{x}right) + 3lnleft(1 + frac{3}{x}right)}{frac{1}{x}}} = e^{limlimits_{x to +infty} frac{frac{1}{x} + frac{4}{x} + frac{9}{x}}{frac{1}{x}}} = e^{-14})答案选 (C)。

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